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Math Help - [SOLVED] f integrable implies f^2 integrable

  1. #1
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    [SOLVED] f integrable implies f^2 integrable

    I'm trying, unsuccessfully at the moment, to show the following, which I know is true: If f is Riemann integrable, then f^2 is Riemann integrable. And, I need to do this without the u,v integrable implies uv integrable. I know there's a way to do it, so if you can point me in the right direction, that would be great! Thanks!
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    MHF Contributor chisigma's Avatar
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    A simple example...

    \int_{0}^{1} \frac{dx}{\sqrt{x}} exists...

    \int_{0}^{1} \frac{dx}{x} doesn't exist...

    Kind regards

    \chi \sigma
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  3. #3
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    f integrable on [0,1] implies f bounded on [0,1], by contraposition, the integral of 1/sqrt{x} does not exist.
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    Can you use that if g(x) is continuous and f(x) integrable then (g\circ f)(x) is integrable?

    If so then just take g(x)=x^2.
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  5. #5
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    Talking

    Yessir, that's the trick. The composition of an integrable function into a continuous function maintains integrability. So, yea, taking your continuous function to be t^2 or whatever gives that f\in R[a,b]\Rightarrow f^2\in R[a,b]
    Cheers!
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  6. #6
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by nqramjets View Post
    f integrable on [0,1] implies f bounded on [0,1], by contraposition, the integral of 1/sqrt{x} does not exist.
    I'm afraid I cant agree with you. Starting from the general integration formula...

    \int x^{a}\cdot dx = \frac{x^{1+a}}{1+a} + c

    ... where a is an arbitrary real [or even complex...] number provided only that a \ne -1, proceeding in trivial way we have...

    \int_{0}^{1} \frac{dx}{\sqrt{x}} = |2\cdot \sqrt{x}|_{0}^{1}= 2

    Kind regards

    \chi \sigma
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    Okay, this may be a slight difference of definition, but here is the issue:
    A function is f:[0,1]\rightarrow \mathbb{R} is Riemann Integrable on [0,1] if there is an L\in \mathbb{R} such that for all \varepsilon >0 there exists \delta >0 such that for every tagged partition \dot{P} of [0,1] with \|\dot{P}\|<\delta, then
    |S(f;\dot{P})-L|<\varepsilon
    Now, if you look at the Riemann sum part you have
    S(f;\dot{P})=\sum_{i=1}^nf(t_i)(x_i-x_{i-1}) but if I choose my tag of the first sub-interval to be 0, ie. t_1=0\in [x_0,x_1] then f(t_i)=\frac{1}{\sqrt{t_i}}=\frac{1}{0} which is undefined. So, f\not\in R[0,1] and therefore we cannot apply the FTC.
    \frac{1}{\sqrt{x}} simply is not a function from [0,1] into \mathbb{R} because 0 is not in the domain.
    What is your definition?
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  8. #8
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    Quote Originally Posted by nqramjets View Post
    Okay, this may be a slight difference of definition, but here is the issue:
    A function is f:[0,1]\rightarrow \mathbb{R} is Riemann Integrable on [0,1] if there is an L\in \mathbb{R} such that for all \varepsilon >0 there exists \delta >0 such that for every tagged partition \dot{P} of [0,1] with \|\dot{P}\|<\delta, then
    |S(f;\dot{P})-L|<\varepsilon
    Now, if you look at the Riemann sum part you have
    S(f;\dot{P})=\sum_{i=1}^nf(t_i)(x_i-x_{i-1}) but if I choose my tag of the first sub-interval to be 0, ie. t_1=0\in [x_0,x_1] then f(t_i)=\frac{1}{\sqrt{t_i}}=\frac{1}{0} which is undefined. So, f\not\in R[0,1] and therefore we cannot apply the FTC.
    \frac{1}{\sqrt{x}} simply is not a function from [0,1] into \mathbb{R} because 0 is not in the domain.
    What is your definition?
    I would also have to agree, since for a function to be Riemann integrable on an interval it has to be bounded. Which \frac{1}{\sqrt{x}} clearly isn't on [0,1]. So even though \int_0^1\frac{dx}{\sqrt{x}} exists, it is actually found by doing the following: \lim_{c\to 0^+}\int_c^1\frac{dx}{\sqrt{x}}.
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  9. #9
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    Exactly, By definition, 1/sqrt{x} is not Riemann integrable, so the theorem that show f-->f^2 does not apply. Hence this does not break the rules. After all, a limit of an integral is a limit of a limiting process, which is where everything gets alittle funny. So, in conclusion, we are both right: f integrable implies f^2 integrable, but improper integrals do not necessarily obey the same rules.
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  10. #10
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    \chi\sigma has confused problems from integration theory with an improper integral from basic calculus.
    In basic calculus, \int_0^1 {\frac{{dx}}{{\sqrt x }}} exist and equals 2.
    But it is also known as an improper integral because \frac{1}{\sqrt x} is not defined at x=0.

    In integration theory the function \frac{1}{\sqrt x} is not Riemann integrable on [0,1], not being bounded there.
    But it does have an improper integral there.
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  11. #11
    Senior Member Sampras's Avatar
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    Quote Originally Posted by Plato View Post
    \chi\sigma has confused problems from integration theory with an improper integral from basic calculus.
    In basic calculus, \int_0^1 {\frac{{dx}}{{\sqrt x }}} exist and equals 2.
    But it is also known as an improper integral because \frac{1}{\sqrt x} is not defined at x=0.

    In integration theory the function \frac{1}{\sqrt x} is not Riemann integrable on [0,1], not being bounded there.
    But it does have an improper integral there.
    I believe it is Lebesgue integral though.
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    Yes, I'm pretty sure it is.
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  13. #13
    MHF Contributor chisigma's Avatar
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    The original Riemann definition is reported here...

    http://en.wikipedia.org/wiki/Riemann_integral

    ... and here we can read...

    Given an f(*) defined in the closed interval [a,b], the finite sequence a = x_{0}<x_{1}<...<x_{n} = b, the finite sequence t_{0},t_{1},...,t_{n-1} subject to the condition x_{i}\le t_{i} \le x_{i+1}, a real number \epsilon >0, if exists a real number \delta>0 so that for any partition \overrightarrow{x},\overrightarrow{t} is...

    |\sum_{i=0}^{n-1} f(t_{i})\cdot (x_{i+1}-x_{i}) -s|<\epsilon (1)

    ... we define...

    s=\int_{a}^{b} f(x)\cdot dx (2)

    Now I have indicated as example the function...

    f(x) = \frac{1}{\sqrt{x}} , 0<x\le 1 (3)

    It is obvious that the (3) is not defined in x=0 , but, as in most of such cases, we can resolve the abiguity defining f(*) [for example...] as follows...

    f(x)= \frac{1}{\sqrt{x}} for 0<x\le 1, =0 for x=0 (4)

    ... so that is...

    f^{2}(x)= \frac{1}{x} for 0<x\le 1, =0 for x=0 (5)

    Now we have to verify if f(*) defined in (4) is Riemann integrable according to (1) and if f^{2}(*) defined in (5) is not Riemann integrable according to (1)...

    Kind regards

    \chi \sigma
    Last edited by chisigma; June 3rd 2009 at 05:02 AM. Reason: prudence... and some trivial mistake...
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    MHF Contributor chisigma's Avatar
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    Now we try to compute the 'Riemann sum' ...

    \sum_{i=0}^{n-1} f(t_{i})\cdot (x_{i+1} - x_{i}) (1)

    ... having as goal to arrive to the integral...

    \int_{0}^{1} \frac{dx}{\sqrt{x}} = 2 (2)

    The most simple choice is...

    x_{i}= \frac {i}{n}

    t_{i}= 0 for i=0 , t_{i}=\frac{i+\alpha}{n} for i=1,2,...,n-1 with 0<\alpha<1 (3)

    Defining...

    f(x)= \frac{1}{\sqrt{x}}= 0 in x=0 (4)

    ... the integral (2) is...

    \int_{0}^{1} \frac{dx}{\sqrt{x}} = lim_{n \rightarrow \infty} \sum_{i=1}^{n-1} \frac {f(t_{i})}{n} = lim_{n\rightarrow \infty} \frac{1}{\sqrt{n}}\cdot \sum _{i=1}^{n-1} \frac{1}{\sqrt{i +\alpha}} (5)

    The limit (5) has been computed in elegant fashion in ...

    http://www.mathhelpforum.com/math-he...-sequence.html

    Kind regards

    \chi \sigma
    Last edited by chisigma; June 3rd 2009 at 05:51 AM.
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  15. #15
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    Obviously if I want to compute the integral of 1/sqrt{x} I can, it can be done is second quarter calculus. The point was that 1/sqrt{x} is NOT Reimann integrable in the strictest, by definition, sense. As a counterexample to f integrable --> f^2 integrable you MUST start with a function that is Reimann integrable. 1/sqrt{x} simply is not on [0,1]. It doesn't matter how you alter the function so that it becomes defined, for it is obviously then not the same function. So the implication is ambiguous.
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